Question

A block of mass M is resting on an inclined plane as shown in the figure. The inclination of the plane to the horizontal is gradually increased. It is found that when the angle of inclination is $\theta$, the block just begins to the slide down the plane. What is the minimum force F applied parallel to the plane that would just make the block move up the plane?

• A. $Mgsin\theta$
• B. $Mgcos\theta$
• C. $2Mgcos\theta$
• D. $2Mgsin\theta$

Answer 1

The correct option is D

$2Mgsin\theta$

The component of weight Mg of the block along the inclined plane is $Mgsin\theta$. The limiting friction is also $Mgsin\theta$. To make the block just move up the plane, the minimum force applied must overcome the component $Mgsin\theta$ of gravitational force as well as frictional force $Mgsin\theta$. Hence, the net force to be applied is $2Mgsin\theta$ and the correct choice is (d).